Android 我一直在犯错误;“url”没有值;
这是我的java代码。在我的android studio中,我的url没有任何价值。我打算从wamp服务器上的sql数据库获取图像Android 我一直在犯错误;“url”没有值;,android,Android,这是我的java代码。在我的android studio中,我的url没有任何价值。我打算从wamp服务器上的sql数据库获取图像 public class MainActivity extends AppCompatActivity implements View.OnClickListener{ private String imagesJSON; private static final String JSON_ARRAY ="result"; private static final
public class MainActivity extends AppCompatActivity implements
View.OnClickListener{
private String imagesJSON;
private static final String JSON_ARRAY ="result";
private static final String IMAGE_URL = "url";
private JSONArray arrayImages= null;
private int TRACK = 0;
private static final String IMAGES_URL =
"http://192.168.43.214/apexStore2/image.php";
private Button buttonFetchImages;
private Button buttonMoveNext;
private Button buttonMovePrevious;
private ImageView imageView;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
imageView = (ImageView) findViewById(R.id.imageView);
buttonFetchImages = (Button) findViewById(R.id.buttonFetchImages);
buttonMoveNext = (Button) findViewById(R.id.buttonNext);
buttonMovePrevious = (Button) findViewById(R.id.buttonPrev);
buttonFetchImages.setOnClickListener(this);
buttonMoveNext.setOnClickListener(this);
buttonMovePrevious.setOnClickListener(this);
}
private void extractJSON(){
try {
JSONObject jsonObject = new JSONObject(imagesJSON);
arrayImages = jsonObject.getJSONArray(JSON_ARRAY);
} catch (JSONException e) {
e.printStackTrace();
}
}
private void showImage(){
try {
JSONObject jsonObject = arrayImages.getJSONObject(TRACK);
getImage(jsonObject.getString(IMAGE_URL));
} catch (JSONException e) {
e.printStackTrace();
}
}
private void moveNext(){
if(TRACK < arrayImages.length()){
TRACK++;
showImage();
}
}
private void movePrevious(){
if(TRACK>0){
TRACK--;
showImage();
}
}
private void getAllImages() {
class GetAllImages extends AsyncTask<String,Void,String>{
ProgressDialog loading;
@Override
protected void onPreExecute() {
super.onPreExecute();
loading = ProgressDialog.show(MainActivity.this, "Fetching
Data...","Please Wait...",true,true);
}
@Override
protected void onPostExecute(String s) {
super.onPostExecute(s);
loading.dismiss();
imagesJSON = s;
extractJSON();
showImage();
}
@Override
protected String doInBackground(String... params) {
String uri = params[0];
BufferedReader bufferedReader = null;
try {
URL url = new URL(uri);
HttpURLConnection con = (HttpURLConnection)
url.openConnection();
con.setRequestProperty("Content-Type",
"application/json;charset=utf-8");
con.setRequestProperty("X-Requested-With",
"XMLHttpRequest");
StringBuilder sb = new StringBuilder();
bufferedReader = new BufferedReader(new
InputStreamReader(con.getInputStream()));
String json;
while((json = bufferedReader.readLine())!= null){
sb.append(json+"\n");
}
return sb.toString().trim();
}catch(Exception e){
return null;
}
}
}
GetAllImages gai = new GetAllImages();
gai.execute(IMAGES_URL);
}
private void getImage(String urlToImage){
class GetImage extends AsyncTask<String,Void,Bitmap>{
ProgressDialog loading;
@Override
protected Bitmap doInBackground(String... params) {
URL url = null;
Bitmap image = null;
String urlToImage = params[4];
try {
url = new URL(urlToImage);
image =
BitmapFactory.decodeStream(url.openConnection().getInputStream());
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
return image;
}
@Override
protected void onPreExecute() {
super.onPreExecute();
loading = ProgressDialog.show(MainActivity.this,"Downloading
Image...","Please wait...",true,true);
}
@Override
protected void onPostExecute(Bitmap bitmap) {
super.onPostExecute(bitmap);
loading.dismiss();
imageView.setImageBitmap(bitmap);
}
}
GetImage gi = new GetImage();
gi.execute(urlToImage);
}
@Override
public void onClick(View v) {
if(v == buttonFetchImages) {
getAllImages();
}
if(v == buttonMoveNext){
moveNext();
}
if(v== buttonMovePrevious){
movePrevious();
}
}
我的php代码从sql数据库获取图像,但是,我的android studio无法获取图像。有什么解决办法吗
<?php
include ('classes/functions.php');
$check_product = "SELECT * FROM products WHERE cat_id = '0';";
$run_product_checking = mysqli_query($con, $check_product);
$result = array();
while($row = mysqli_fetch_array($run_product_checking)){
array_push($result,
array('product_img1'=>$row[4]
));
}
echo json_encode(array("result"=>$result));
?>
将IP地址更改为其域名值。根据您的AsyncTask类,您将url作为第一个参数传递
class GetImage extends AsyncTask<String,Void,Bitmap>{
.....
}
String urlToImage = params[4];
要解决此问题,只需将参数值从4替换为0
String urlToImage = params[0];
PS:确保JSON字符串“IMAGE_URL”不返回空值
getImage(jsonObject.getString(IMAGE_URL));
为了避免从JSON中获取空值,您可以使用jsonObject.optString(IMAGE_URL)来处理空指针。无法更改参数[4],因为将给出java.lang.ArrayIndexOutOfBoundsException:length=1;index=4错误。顺便说一下,我得到的错误SkImageDecoder::Factory返回null。如何更改代码?能否在“String urlToImage=params[4]”行后打印“urlToImage”值,并告诉我结果如何?
getImage(jsonObject.getString(IMAGE_URL));